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Math Help - Maclaurin series for Ln Z! (( Gamma Function ))

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    Unhappy Maclaurin series for Ln Z! (( Gamma Function ))

    Last edited by mr fantastic; November 4th 2010 at 04:55 PM. Reason: Begging in title.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by amro05 View Post
    What is F? Is \text{Ln} the principal branch of the logarithm?
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    Quote Originally Posted by Drexel28 View Post
    What is F? Is \text{Ln} the principal branch of the logarithm?
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by amro05 View Post
    The 'factorial function' is defined as...

    \displaystyle z! = \int_{0}^{\infty} t^{z}\ e^{-t}\ dt = \Gamma(z+1)= z\ \Gamma(z) (1)

    ... and Your request [if I undestood correctly...] is to find the Mc Laurin series of \ln z!. Very well!... let's start from the well known 'inifinite product'...

    \displaystyle \frac{1}{\Gamma (z)} = z\ e^{\gamma\ z}\ \prod_{n=1}^{\infty} (1+\frac{z}{n})\ e^{-\frac{z}{n}} (2)

    ... where...

    \displaystyle \gamma = \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{1}{k} - \ln n (3)

    ... is the so called 'Euler's constant'. From (1) and (2) we derive...

    \displaystyle \frac{1}{z!} =  e^{\gamma\ z}\ \prod_{n=1}^{\infty} (1+\frac{z}{n})\ e^{-\frac{z}{n}} (4)

    ... and from (4)...

    \displaystyle \lambda(z)= \ln z! = -\gamma\ z - \sum_{n=1}^{\infty} \{\ln (1+\frac{z}{n}) - \frac{z}{n} \} (5)

    Deriving both terms of (5) we obtain...

    \displaystyle \lambda^{'}(z) = -\gamma - \sum_{n=1}^{\infty} \{\frac{1}{n}\ (1+\frac{z}{n})^{-1} - \frac{1}{n} \} (6)

    ... so that is...

    \displaystyle \lambda^{'}(0) = -\gamma (7)

    Deriving both terms of (6) we obtain...

    \displaystyle \lambda^{''}(z) = \sum_{n=1}^{\infty} \frac{1}{n^{2}}\ (1+\frac{z}{n})^{-2} (8)

    ... so that is...

    \displaystyle \lambda^{''}(0) = \zeta (2) (9)

    ... where \displaystyle \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}} is the so called 'Riemann zeta function'. Proceeding along the way we find for n>1 ...

    \displaystyle \lambda^{(n)}(0) = (-1)^{n}\ (n-1)!\ \zeta(n) (10)

    ... so that the requested McLaurin expansion is...

    \displaystyle \ln z! = -\gamma\ z + \sum_{n=2}^{\infty} (-1)^{n}\ \zeta(n)\ \frac{z^{n}}{n} (11)

    The (11) converges for |z|<1...

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 5th 2010 at 02:25 PM. Reason: Corrected formulas (5) and (6)...
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    Many many thanks

    all the best to you




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